我在加载页面上有一个带有花式框的弹出窗口.
I have a popup with fancybox that appear at load page.
如果用户更改页面并返回带有弹出窗口的页面没有第二次显示,我需要显示一次弹出窗口.
I need to show the popup once a time, if the user change page and back on the page with popup doesn't reveal a second time.
我读过可以使用 cookie 插件 (https://github.com/carhartl/jquery-cookie)但我不明白如何在这段代码中集成......
I've read that could be use a cookie plug in (https://github.com/carhartl/jquery-cookie) but i dont understant how integrate in this code...
我有一个简单的 html/css 网站.
I have a simple site in html/css.
这是代码:
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title></title>
<script type="text/javascript" src="http://code.jquery.com/jquery-1.4.2.min.js"> </script>
<script type="text/javascript" src="jquery.fancybox-1.3.1.js"></script>
<link rel="stylesheet" type="text/css" href="jquery.fancybox-1.3.1.css" media="screen" />
<script src="jquery.cookie.js"></script>
<script type="text/javascript">
function openFancybox() {
setTimeout(function () {
$('#yt').trigger('click');
}, 500);
};
$(document).ready(function () {
var visited = $.cookie('visited');
if (visited == 'yes') {
return false; // second page load, cookie active
} else {
openFancybox(); // first page load, launch fancybox
}
$.cookie('visited', 'yes', {
expires: 7 // the number of days cookie will be effective
});
$("#yt").click(function() {
$.fancybox({
'padding' : 0,
'autoScale' : false,
'transitionIn' : 'none',
'transitionOut' : 'none',
'title' : this.title,
'width' : 680,
'height' : 495,
'href' : this.href.replace(new RegExp("watch\?v=", "i"), 'v/'),
'type' : 'swf',
'swf' : {
'wmode' : 'transparent',
'allowfullscreen' : 'true'
}
});
return false;
});
});
</script>
</head>
<body onload='$("#yt").trigger("click");'>
<a id="yt" href="https://www.youtube.com/watch?v=ROTYmNckBCw&fs=1&autoplay=1"><img src="http://fancyapps.com/fancybox/demo/1_s.jpg" alt=""/></a>
</body>
</html>
为了浏览器的一致性,第一次可能需要延迟fancybox加载执行,试试这段代码:
For browser consistency, you may need to delay the fancybox load execution for the first time so try this code :
function openFancybox() {
// launches fancybox after half second when called
setTimeout(function () {
$('#yt').trigger('click');
}, 500);
};
$(document).ready(function () {
var visited = $.cookie('visited'); // create the cookie
if (visited == 'yes') {
return false; // second page load, cookie is active so do nothing
} else {
openFancybox(); // first page load, launch fancybox
};
// assign cookie's value and expiration time
$.cookie('visited', 'yes', {
expires: 7 // the number of days the cookie will be effective
});
// your normal fancybox script
$("#yt").click(function () {
$.fancybox({
// your fancybox API options
});
return false;
});
});
查看这里的代码 JSFIDDLE
See code at this JSFIDDLE
注意事项:
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